# Find the point $D$ where the bisector of $\angle A$ meets side $BC$ of the triangle $ABC$

Consider a triangle $$ABC$$. Let the angle bisector of angle $$A$$ intersect side $$BC$$ at a point $$D$$ between $$B$$ and $$C$$. The angle bisector theorem states that the ratio of the length of the line segment $$BD$$ to the length of segment $$DC$$ is equal to the ratio of the length of side AB to the length of side $$AC$$:

$${\displaystyle {\dfrac {|BD|}{|DC|}}={\frac {|AB|}{|AC|}},} {{\dfrac {|BD|}{|DC|}}}={{\dfrac {|AB|}{|AC|}}}$$,The vector $$A,B,C$$ of triangle $$ABC$$ have respectively position vectors $$\vec{a}, \vec{b}, \vec{c}$$ with respect to a given origin $$O$$. Show that the point $$D$$ where the bisector of $$\angle A$$ meets $$BC$$ has position vector $$\vec{d}=\frac{\beta \vec b+\gamma \vec c}{\beta+\gamma}$$, where $$\beta=|\vec c-\vec a|$$ and $$\gamma=|\vec a-\vec b|$$. Hence deduce that the incentre $$I$$ has position vector $$\frac{\alpha \vec a+\beta \vec b+\gamma \vec c}{\alpha+\beta+\gamma}$$, where $$\alpha=|\vec b-\vec c|$$

We can prove this using the result "The angle of bisector of the angle $$\angle A$$" of a triangle $$ABC$$ divides the sides $$BC$$ in the ration $$AB:AC$$.

Is there any other way to prove this result?

Since the angle bisector of $\widehat{BAC}$ is the locus of points $P$ for which $d(P,AB)=d(P,AC)$, the trilinear coordinates of the incenter $I$ are $[1;1;1]$. It follows that the barycentric coordinates of the incenter are $[a;b;c]$, hence the exact barycentric coordinates are $\left[\frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}\right]$, i.e.

$$I = \frac{aA+bB+cC}{a+b+c}$$ as wanted.

A simple (although possibly tedious) approach is to prove it algebraically. I will simply write the lowercase letters for the vectors.

From the angle equality

$$\frac{(d-a)\cdot (c-a)}{|c-a|}=\frac{(b-a) \cdot (d-a)}{|b-a|}$$

Since $d-c$ and $b-d$ are collinear,

$$d-c = \delta(b-d)$$ $$d = \frac{\delta b+ c}{1+\delta}$$ (this is a general result for a point lying on a line determined by two position vectors)

$$d-a = \frac{\delta (b-a)+(c-a)}{1+\delta}$$

$$\frac{(d-a).(c-a)}{|c-a|} = \frac{\delta (b-a)\cdot (c-a)}{(1+\delta)|c-a|}+\frac{|c-a|}{1+\delta}$$

$$\frac{(b-a).(d-a)}{|b-a|} = \frac{\delta |b-a|}{(1+\delta)}+\frac{(c-a)\cdot (b-a)}{(1+\delta)|b-a|}$$

Equate and solve for $\delta$ to get the first part.

For the second part, we use a similar approach. For any two angle bisectors, the incentre will have the general form:

$$\frac{a+\mu d}{1+\mu} = \frac{b+\eta e}{1+\eta}$$

We get two equations to solve for two variables. This will give the result.